(1) First we split the reactions into two half reactions: Cd ---> Cd 2+ PbO 2 --
ID: 700033 • Letter: #
Question
(1)
First we split the reactions into two half reactions:
Cd ---> Cd2+
PbO2 ---> Pb2+
Next we balance all atoms other than H and O:
Cd ---> Cd2+
PbO2 ---> Pb2+
Next we balance O by adding H2O molecules.
Cd ---> Cd2+
PbO2 ---> Pb2++ 2H2O
Next we balance H by adding protons:
Cd ---> Cd2+
PbO2 + 4H+ ---> Pb2++ 2H2O
Next we balance charge by adding electrons:
Cd ---> Cd2+ + 2e-
PbO2 + 4H+ + 2e- ---> Pb2++ 2H2O
Next we add the two equations and cancel the common terms.
PbO2 + 4H+ + Cd ---> Pb2++ 2H2O + Cd2+
Hope this helps !
Explanation / Answer
When the following equation is balanced properly under acidic conditions, what are the coefficients of the species shown? Cd + Pbo Water appears in the balanced equation as a(reactant, product, neither) with a coefficient of (Enter 0 for neither.) How many electrons are transferred in this reaction? Submit Answer Retry Entire Group 9 more group attempts remaining
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